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This may be a trivial question, but I am not sure how current flows in certain types of transmission lines. Consider for example a coaxial cable working in its TEM mode. The wave travels perpendicular to the plane, and the fields are contained in the plane.

Structure of the electric and magnetic fields within coaxial line

Source: Structure of the electric and magnetic fields within coaxial line. In this case, the wave is propagating away from the viewer by Ellingson, Steven W. (2018) Electromagnetics, Vol. 1. Blacksburg, VA: VT Publishing. https://doi.org/10.21061/electromagnetics-vol-1 CC BY-SA 4.0

From the treatment of general transmission lines (working with V and I, instead of the electric and magnetic fields), we know the current propagates in what would be the direction perpendicular to the plane (+z and/or -z direction). This is where my question arises. If the current density is proportional to the electric field how can the current flow perpendicular to the plane if the electric field is radial?

I have the same problem when dealing with other types of transmission lines such as microstrip and stripline, but understanding the coaxial cable first may help me understand the other cases.

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  • \$\begingroup\$ If you are happy with the direction and shape of the B field (in blue) then current flow has to be in/out of the page. One follows from the other. \$\endgroup\$
    – Andy aka
    Aug 28, 2023 at 8:36
  • \$\begingroup\$ @Andyaka yes I get that, but how is that possible if this is TEM mode? If carrent flows in the perpendicular direction, then there is E field in the direction of propagation. Therefore, it wouldn’t be a TEM mode. \$\endgroup\$ Aug 28, 2023 at 11:59
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    \$\begingroup\$ There will be a small E field that allows the current, the loss resistance and the E field to maintain ohm's law. \$\endgroup\$
    – Andy aka
    Aug 28, 2023 at 15:47
  • \$\begingroup\$ @Andyaka could you elaborate more? Books say the mode is TEM, and do not mention that there is an electric field in the direction of propagation (otherwise the mode would not be TEM). Furthermore, some books obtain an expression for the current density from the magnetic field, but I do not know how this does not produce an electric field tangential to the conductor. \$\endgroup\$ Aug 28, 2023 at 17:39
  • \$\begingroup\$ Not really; if I could I would give a formal answer. \$\endgroup\$
    – Andy aka
    Aug 28, 2023 at 18:33

2 Answers 2

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It's perfectly valid physics to have a coaxial line with perfect conductors. It supports a TEM mode, this operates all the way down to DC. If your picture represents DC power flowing along the cable then there are surface currents on the outside of the inner conductor going into the page, and on the inside of the outer conductor coming out of the page. If the conductors are perfect then there is no loss.

The currents on the conductors are such that the boundary conditions of the fields are satisfied - so currents flow consistent with the tangential B.

A perfect conductor has a tangential E field = 0 but you still get currents flowing. This happens when you set up some external fields and the charges on the conductor rearrange themselves to maintain the conductor at a constant potential. If the external fields are AC then the charges continually rearrange themselves to make \$E_t = 0\$, so alternating currents flow.

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  • \$\begingroup\$ Excellent answer. It's exactly what I have later seen in the books. Thank you! \$\endgroup\$ Sep 12, 2023 at 16:16
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When current density is proportional to the electric field, the ratio is the resistance: since there is a large electric field between the inner and outer conductors of your example, there would be a large current if the insulation failed.

And, to the extent that current exists in the axial direction, there must also be an axial electric field. That doesn't show up on your diagram because it is perpendicular to your diagram.

But at Radio Frequencies most of the energy transfer is not happening by electron flow (remember that you can get radio transmission in free space), and the energy required to heat up the inner and outer conductors is not what you are typically interested in, so (using the principle of superposition) the axial component of the electric field doesn't need to be analyzed.

From further comments made, it appears that the questioner is only interested in the absence of axial electric field in simplified examples. That has a simplified answer: when the conductor is infinitely conductive, the E field required to excite current is infinitely small.

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  • \$\begingroup\$ Ok but if there is an electric field component in the direction of propagation, why is this a TEM mode? By definition it would not be. \$\endgroup\$ Aug 28, 2023 at 12:01
  • \$\begingroup\$ The TEM mode is not associated with electron flow. There would be electron flow if the insulation failed. To the extent that there is current flow perpendicular to the plane, it is not a simple TEM field. Where there is axial current flow, there is also a non-TEM mode. Wave guides and Transmission lines are not really lossless: other modes are excited. Your diagram doesn't even show the field in the conductor: you show the field in the insulator and then ask about modes in the conductors. \$\endgroup\$
    – david
    Aug 29, 2023 at 4:59
  • \$\begingroup\$ Modes in the conductor? Assuming this is a perfect electric conductor the fields are zero inside it. Also, I do not understand your argument linking losses and excitation of other modes. Excitation of modes depends on frequency, all books I have read analyze losses working with a particular mode. Anyway, I have looked up the explanation in some books and the current density is derived from the boundary condition of the magnetic field. \$\endgroup\$ Sep 11, 2023 at 3:10
  • \$\begingroup\$ If the conductor is infinitely conductive, infinite current will be excited by an infinitely small E field. If you are actually interested in the actual (real, physical) current flowing in the conductor, you need to read a different kind of book. Certainly any book that assumes a "perfect electric conductor' is the wrong book. \$\endgroup\$
    – david
    Sep 11, 2023 at 8:39
  • \$\begingroup\$ What book do you recommend that has this information? \$\endgroup\$ Sep 11, 2023 at 15:15

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